A uniform disc of mass m and radius r is pivoted at point P and is free to rotate in vertical plane . The centre C of disc is initially in horizontal position with P .If it is released from this position , then its angular acceleration when the line PC is inclined to the horizontal at angle Q ?

Moment of inertia of a disc about its centre is12mr2
When the disc makes an angle Q with the horizontal

Consider the distance between P and C to be x.
Then the moment of inertia about P is
I=I0+mx2I=12mr2+mx2
Hence
mgcosQx=Iαα=2mgxcosQmr2+2x2α=2gxcosQr2+2x2

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